# 3-dimensional potential energy problem

1. Oct 25, 2004

### winhog

I have a problem on my homework that says the potential energy of a particle is given by its position in the x-y plane according to

P.E. = x^3 + 8x^2 + 34yz

and I have to calculate the force on the particle at point (x,y,z), and all equilibrium points.

dU = 3x^2 dx + 16x dx + 34y dz + 34z dy

and F = -dU/dr with r being the vector position...

-F = 3x^2 dx/dr + 16x dx/dr + 34y dz/dr + 34z dy/dr

but that can't be the answer can it? I'm not given any dx/dr's or anything like that...I have a feeling there's some simple math thing I'm missing.

Any hints anyone can think of?

2. Oct 25, 2004

### Tide

You need to find the gradient of the potential - it's a vector quantity!

3. Oct 25, 2004

### winhog

Hmmm...I think my problem might be I've never learned how to take a vector derivative and this might be above my head mathematically

Maybe I can separate the potential energy into x and y components...but i really don't know how.

I guess since the particle is on the x-y plane, dz = 0...so my equation can be simplified to

F = (-3x^2 - 16x) dx/dr - 34z dy/dr .

Can I change dx/dr and dy/dr into (x-direction) and (y-direction) in terms of the force? Then use pythagorean theorem to find the total force? Am I just rambling?

4. Oct 25, 2004

### Tide

It's really straightforward:

The x-component of the force is the partial derivative of the potential with respect to x. For the y-component use the partial wrt y and for the z-component use the partial wrt z. Voila!

5. Oct 25, 2004

### nrqed

One caveat: the force is MINUS the gradient of the potential, $\vec F = - \vec \nabla V$

or, to be more specific,

$F_x = - {\partial V(x,y,z) \over \partial x}$

$F_y = - {\partial V(x,y,z) \over \partial y}$

$F_z = - {\partial V(x,y,z) \over \partial z}$

If the particle is indeed constrained to the xy plane, then the z component of the force will be canceled by a normal force. In the x and y component, one must then set the value of z corresponding to the z plane one is in (z=0 or some other value).

Pat

6. Oct 25, 2004

### Tide

nrqed,

Thanks - I meant to type "proportional to" but somehow it didn't come out!

7. Oct 26, 2004

### winhog

If I take the derivative with respect to, say, x, can I assume y and z are constants? If so, I messed up on a pretty simple problem

8. Oct 26, 2004

### Tide

That's essentially what you do when you take partial derivatives which applies here.

9. Oct 27, 2004

### winhog

Thanks for the help guys!